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Microeconomics C, Final Exam, February 2011

Suggested Answers

March 1, 2011

1. (a) First, B1 is strictly dominated by A1. Then A2 and B2 are strictlydominated by C2. Then A1 is strictly dominated by C1. Finally, D2

is strictly dominated by C2. So the solution is

(C1; C2):

(b) Pure NE:(T; L) and (B; R):

To nd the Mixed NE, let p denote the probability that player 1 playsT and let q denote the probability that player 2 plays L. In a mixedNE with p; q2 (0; 1) each player must be indierent between his purestrategies:

q+ (1 q) = 3(1 q)

p + (1p) = 2(1p)

From these equations we get the only (non-pure) MNE:

(p; q) = (1

2;

2

3):

(c) i. There is one subgame - it starts at player 3s decision node. Theset of strategies for player 1 is:

fL;Rg:

The set of strategies for player 2 is:

fL0; R0g:

The set of strategies for player 3 is:

fL00; R00g:

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ii. The only pure strategy SPNE is:

(R;L0; R00):

To show this, rst note that in a SPNE player 3 must take theoptimal action at his decision node. I.e., he must play R00. Player1 and 2 will realize this and thus they will take into account thatif they play L and L0 then their payos will be 1 (for player 1)and 6 (for player 2). Therefore player 1 and 2 are really playingthe following game:

L0 R0

L 1; 6 2; 5R 4; 3 0; 2

The only pure NE of this game is (R;L0) and thus the only pure

SPNE of the entire game is (R;L0; R00

).iii. The pure strategy NE of the game are:

(R;L0; R00); (R;L0; L00); (L;R0; L00):

From question ii we know that only the rst of these is subgameperfect. One way to nd all the pure strategy NE is to writethe normal form as two matrices (player 1 cooses a row, player2 chooses a column, player 3 chooses a matrix):

L0 R0

L 4; 3; 2 2

; 5

; 4

R 4

; 3

; 1

0; 2; 0

L00

L0 R0

L 1; 6; 3

2; 5; 4

R 4

; 3

; 1

0; 2; 0

R00

Of course it is sucient to nd one NE that is not subgameperfect.

2. (a) Firm i solves the problem

maxqi0

(70 (qi + qj))qi (qi)2:

First order condition:

70 qj 4qi = 0:

Best response fct:

qi = 70 qj4:

Pure strategy NE:q1

= q2

= 14:

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(b) Firm 1 solves:

maxq10

(70 (q1 + 70 q14

))q1 (q1)2:

FOC:3

470

14

4q1 = 0 ) q1 =

3

1470 = 15:

Quantity of rm 2 then is

70 15

4=

55

4:

So the outcome of the SPNE is:

(q1; q2) = (15;55

4):

Formally, the SPNE is(15;

70 q14

):

(c) Since 554< 14 < 15 we have

q1> q1 and q

2< q2:

So Firm 1 produces more in the sequential game, Firm 2 producesless. When Firm 1 moves rst it can commit to a larger quantitywhich makes Firm 2 choose a smaller quantity.

(d) The merged entity solves the following problem:

maxqm0

(70 qm)qm (1 s)(qm)2:

The rst order condition is:

70 2(2 s)qm = 0 ) qm =70

2(2 s):

Note that the quantity is increasing in s. This is also intuitivelyobvious since a higher s decreases the marginal cost of the mergedentity.

In (a) the aggregate quantity in equilibrium was 28. So we have tond s such that

70

2(2 s)= 28 ) (2 s) =

35

28=

5

4) s = 2

5

4=

3

4:

3. (a) The set of feasible payos is the convex hull of the four payo pairsin the matrix. I.e., the line pieces connecting the points

f(2; 2); (4; 1); (0; 4); (3; 3)g

and all points "inside" these line pieces (see p. 96 in Gibbons). Seegure below.

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(b) First note that the unique NE ofG is (T; L) which gives the players apayo of 2 each. It then follows from Friedmans Theorem (Gibbonsp. 97) that the set of feasible payos that can be achieved as averagepayos in a SPNE of G(1; ) is the set of feasible payos (u1; u2)with u1; u2 > 2. See gure below.

4. (a) [(L;L); (d; d); p = 12; q] is a PBE for q 1

2. These are the only PBEs

where the senders strategy is (L;L).

(b) R is a dominated message for type t2 (but not for type t1). Therefore,

none of the PBE from (a) satisfy Signalling Requirement 5 (Gibbonsp. 237). Because that would require the receivers belief after ob-serving R to put no probability mass on the sender being type t2,i.e., we should have q = 1. And this is not the case in any of thePBE from (a).

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